NO
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


Termination w.r.t. Q of the given QTRS could be disproven:

(0) QTRS
(1) NonTerminationProof [COMPLETE, 913 ms]
(2) NO


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(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   b(b(x1)) -> b(a(x1))
   a(b(x1)) -> b(c(c(x1)))
   a(c(x1)) -> a(a(x1))
   c(a(x1)) -> b(a(a(x1)))

Q is empty.

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(1) NonTerminationProof (COMPLETE)
We used the non-termination processor [OPPELT08] to show that the SRS problem is infinite.

Found the self-embedding DerivationStructure: 
"b a (c)^k b a -> b a (c)^2k+1 b a a"
b a (c)^k b a -> b a (c)^2k+1 b a a
by Overlap u with l (ol4)"a (c)^k b a -> b (c)^2k+1 b a a
by Equivalent"a (c)^k b a -> b (c c)^k c b a a
by Overlapping Derivationstructures"a (c)^k b a -> (a)^k b c b a a
by Overlap u with r (ol3)"a (c)^k -> (a)^k a
by Selfoverlapping OC am2"a c -> a a
by original rule (OC 1)"""a b a -> b c b a a
by OverlapClosure OC 2"a b -> b c c
by original rule (OC 1)""c a -> b a a
by original rule (OC 1)""""(a)^k b -> b (c c)^k
by Selfoverlapping OC am1"a b -> b c c
by original rule (OC 1)"""""b b -> b a
by original rule (OC 1)"

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(2)
NO